How much heat is required in order to heat 30 g of alcohol taken at a temperature of 28 degrees Celsius to a boil and turn into steam?

Data: m (weight of alcohol taken) = 30 g (0.03 kg); t (temperature at which alcohol was taken) = 28 ºС.

Reference data: С (specific heat capacity) = 2500 J / (kg * ºС); tboil (temperature of vaporization of ethyl alcohol) = 78.3 ºС (≈ 78 ºС); L (specific heat of vaporization) = 9 * 10 ^ 5 J / kg.

1) Heating alcohol to boiling: Q1 = C * m * (tboil – t) = 2500 * 0.03 * (78 – 28) = 3750 J.

2) Conversion of alcohol to steam: Q2 = L * m = 9 * 10 ^ 5 * 0.03 = 27000 J.

3) Required heat: Q = 3750 + 27000 = 30750 J.